If y = frac{x^4 - x^2 + 1}{x^2 + sqrt{3}x + 1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $y = \frac{x^4 - x^2 + 1}{x^2 + \sqrt{3}x + 1}$.We can rewrite the numerator as $x^4 + 2x^2 + 1 - 3x^2 = (x^2 + 1)^2 - (\sqrt{3}x)^2$.Using

Vedclass · Accepted Answer

$\cot\frac{5\pi}{12}$

Vedclass · Answer

(B) Given $y = \frac{x^4 - x^2 + 1}{x^2 + \sqrt{3}x + 1}$.We can rewrite the numerator as $x^4 + 2x^2 + 1 - 3x^2 = (x^2 + 1)^2 - (\sqrt{3}x)^2$.Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $y = \frac{(x^2 + 1 - \sqrt{3}x)(x^2 + 1 + \sqrt{3}x)}{x^2 + \sqrt{3}x + 1}$.Simplifying the expression,we have $y = x^2 - \sqrt{3}x + 1$.Now,differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x - \sqrt{3}$.Comparing this with $\frac{dy}{dx} = ax + b$,we find $a = 2$ and $b = -\sqrt{3}$.Thus,$a + b = 2 - \sqrt{3}$.Since $	an(15^\circ) = 	an(\frac{\pi}{12}) = 2 - \sqrt{3}$,and $	an(\frac{\pi}{12}) = \cot(\frac{\pi}{2} - \frac{\pi}{12}) = \cot(\frac{5\pi}{12})$.Therefore,$a + b = \cot\frac{5\pi}{12}$.

$I$. $f'(x) = \frac{3}{4}x^{4/3} + 9$	$II$. $f(x) = \frac{9}{28}x^{7/3} - 2$
$III$. $f(x) = \frac{9}{28}x^{7/3} + 6$	$IV$. $f'(x) = \frac{3}{4}x^{4/3} - 4$

If $y = \frac{x^4 - x^2 + 1}{x^2 + \sqrt{3}x + 1}$ and $\frac{dy}{dx} = ax + b$,then the value of $a + b$ is equal to

Similar Questions

Differentiate the function with respect to $x$: $\cos (\sin x)$

Which of the following statements could be true if $f''(x) = x^{1/3}$?

$I$. $f'(x) = \frac{3}{4}x^{4/3} + 9$ $II$. $f(x) = \frac{9}{28}x^{7/3} - 2$

$III$. $f(x) = \frac{9}{28}x^{7/3} + 6$ $IV$. $f'(x) = \frac{3}{4}x^{4/3} - 4$

If $f^{\prime}(x)=a \cos x+b \sin x$ and $f^{\prime}(0)=4, f(0)=3, f\left(\frac{\pi}{2}\right)=5$,then $f(x)=$

Let $f : R \rightarrow R$ be defined as $f(x) = x^{3} + x - 5$. If $g(x)$ is a function such that $f(g(x)) = x, \forall x \in R$,then $g^{\prime}(63)$ is equal to

If $f(2)=4$ and $f^{\prime}(2)=1$,then $\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module